Properties of Eigenvalues

Eigenvalues are the special set of scalar values that is associated with the set of linear equations most probably in the matrix equations. The eigenvectors are also termed as characteristic roots. It is a non-zero vector that can be changed at most by its scalar factor after the application of linear transformations.

What Are Eigenvalues?

Eigenvalues describes the relationship between its coefficients and vectors of a square matrix. The linear transformation of eigen values is function that maps vectors from one vector space to another space such that vector addition and scale multiplication remain unchanged.

In mathematical terms, consider square matrix (A) (which stands for a linear transformation), and the eigenvalue (λ) of an eigenvector (v) is a scalar such that:

\bold{Av=\lambda v}

Where:

• A is an n × n square matrix.
• v is the eigenvector corresponding to λ, and it is a non-zero vector.
• λ is the eigenvalue, a scalar.

The equation suggests us that when the matrix A acts on the eigenvector v then the result is simply the eigenvector scaled by the eigenvalue λ.

How to Find Eigenvalues?

To find eigenvalues, one must solve the characteristic equation. The characteristic equation is derived from the determinant of the matrix A subtracted by λ times the identity matrix I:

det (A − λI) = 0

Solving this equation gives the eigenvalues of the matrix A.

Example: The matrix A is given as: A = \begin{pmatrix}2 && 0 \\ 0 && 3\end{pmatrix}. Now, find the eigenvalues of the given matrix.

Solution:

By solving the characteristic equation, we can find the eigenvalues. The characteristic equation is given as: Av=\lambda v.

\therefore A - \lambda I = \begin{pmatrix}2 && 0 \\ 0 && 3\end{pmatrix} - \lambda \begin{pmatrix}1 && 0 \\ 0 && 1\end{pmatrix} = \begin{pmatrix}2-\lambda && 0 \\ 0 && 3-\lambda \end{pmatrix}

Now, calculating the determinant of the matrix (A - λI)

\therefore det(A - \lambda I) = (2 - \lambda)(3 - \lambda)

Solving the equation for getting the eigenvalues

∴ (2 - λ) (3 - λ) = 0

∴ λ₁ = 2 and λ₂ = 3

Hence, the eigenvalues of the matrix given are λ₁ = 2 and λ₂ = 3.

Properties of Eigenvalues

There are many such properties of eigenvalues which are mentioned below.

Sum of Eigenvalues

A matrix A's eigenvalues sum of the eigenvalues is the same as the matrix' trace. If λ₁, λ₂, . . ., λ_n are the eigenvalues of the matrix A, then

\sum_{i=1}^n \lambda_i = \text{tr}(A)

Note: The trace of a matrix A, denoted as tr(A), is the sum of the elements on its main diagonal.

Product of Eigenvalues

A matrix A's eigenvalues product of the eigenvalues is the same as the matrix determinant. For an n × n matrix A, if λ₁, λ₂, . . ., λ_n​ are the eigenvalues of A, then:

Product of Eigenvalues = λ₁ ⋅ λ₂ ⋅ . . . ⋅ λ_n = det(A)

Explanation: The determinant of a matrix is a scalar value that provides useful information about the matrix, including properties related to its eigenvalues. The determinant of a matrix can be calculated by multiplying all the eigenvalues of the matrix together.

Eigenvalues of a Diagonal Matrix

On a diagonal matrix D, one obtains the eigenvalues by reading off the diagonal entries. Let's say:

\therefore D = \begin{pmatrix} d_1 & 0 & \cdots & 0 \\ 0 & d_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_n \end{pmatrix}

On solving the following we will get the eigenvalues:

\therefore \text{det}(D - \lambda I) = \text{det} \begin{pmatrix} d_1 - \lambda & 0 & \cdots & 0 \\ 0 & d_2 - \lambda & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_n - \lambda \end{pmatrix} = (d_1 - \lambda)(d_2 - \lambda) \cdots (d_n - \lambda)

This determinant when set to '0' the resultant we get is the eigenvalues:

λ_i = d_i for i = 1, 2,........., n

This feature is the most useful for a number of transformations and computations of matrices, as diagonal matrices are the easiest to use due to their clear eigenvalue structure. In addition, it is a practical application of diagonalization that allows simplification of various complex matrix operations.

Eigenvalues of Triangular Matrices

Eigenvalues of each arrangement, whether upper triangular or lower triangular matrices, are the diagonal elements of them.

Let T be an upper triangular matrix:

\therefore T = \begin{pmatrix} t_{11} & t_{12} & \cdots & t_{1n} \\ 0 & t_{22} & \cdots & t_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & t_{nn} \end{pmatrix}

On solving the following we will get the eigenvalues:

\therefore \text{det}(T - \lambda I) = \text{det} \begin{pmatrix} t_{11} - \lambda & t_{12} & \cdots & t_{1n} \\ 0 & t_{22} - \lambda & \cdots & t_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & t_{nn} - \lambda \end{pmatrix} = (t_{11} - \lambda)(t_{22} - \lambda) \cdots (t_{nn} - \lambda)

The determinant simply results to the product of the diagonal terms, so the eigenvalues are:

λ_i = t_ii for i = 1, 2,......., n

This rule greatly simplifies the Eigenvalue finding process in the context of triangular matrices, which often tend to show up in numeric algorithms such as LU decomposition.

Eigenvalues of the Inverse Matrix

If λ is the eigenvalue of the matrix A, then \frac{1}{λ} is an eigenvalue of A^-1.

Let's say: Av = vλ by doing A^-1 to both sides:

\therefore A^{-1}A\mathbf{v} = A^{-1}(\lambda \mathbf{v})

Which becomes: \mathbf{v} = \frac{1}{\lambda} A^{-1} \mathbf{v}

Thus: A^{-1}\mathbf{v} = \frac{1}{\lambda}\mathbf{v}

This means that the reciprocal of \frac{1}{\lambda} is an eigenvalue of A^-1. This property is particularly useful in solving systems of linear equations and understanding the behavior of matrices under inversion.

Eigenvalues of a Scalar Multiple of a Matrix

If ƛ is an eigenvalue of a matrix A, then cλ is an eigenvalue of cA, where c is a scalar. Let us take Av = λv, then:

(cA) v = c (Av) = c (λv) = cλv

Mathematically, cλ is an eigenvalue of cA.

Eigenvalues of the Transpose

The eigenvalues of a matrix A and its transpose A^T are identical. This is because the characteristic polynomial of A is the same as that of A^T.

Mathematicall,

det(A - λ I) = det(A^T - λ I)

Therefore, A and A^T possess the same set of eigenvalues.

Eigenvalues and Matrix Powers

If λ is the eigenvalue of the matrix A, then λ^k is an eigenvalue of A^k, where k is a positive integer.

Suppose: Av = λv, then applying A multiple times:

A^k\mathbf{v} = A^{k-1}(A\mathbf{v}) = A^{k-1}(\lambda\mathbf{v}) = \lambda A^{k-1}\mathbf{v} = \lambda^2 A^{k-2}\mathbf{v} = \cdots = \lambda^k \mathbf{v}

This shows that ƛ^k is an eigenvalue of A^k. This property is useful in understanding the long-term behavior of the dynamic systems described by powers of matrices.

Eigenvalues of a Hermitian Matrix

All of the eigenvalues are real for a Hermitian matrix (i.e. A^H = A, where A^H is the conjugate transpose).

This is a consequence of the fact that such a Hermitian matrix can be diagonalized by an unitary transformations and hence 2. Now, the λ in A will be real for any eigenvalue of it if and only if A is Hermitian mathematically.

Eigenvalues of a Skew-Hermitian Matrix

For a skew-Hermitian matrix A (i.e., A^H = -A), all the eigenvalues are purely imaginary or zero otherwise.

This property can be derived from the fact that for a skew-Hermitian matrix: λ is purely imaginary or zero for any eigenvalue of A

If (v) is an eigenvector of (A) with eigenvalue ( λ ), then: Av = λv

Taking the Hermitian transpose:

\mathbf{v}^H A^H = \lambda^* \mathbf{v}^H

Since A^H = -A :

\mathbf{v}^H (-A) = \lambda^* \mathbf{v}^H

leading to: -λ = λ*

implying λ is purely imaginary.

Eigenvalues of an Orthogonal Matrix

If A is an orthogonal matrix (i.e., A^T A = I ), then all eigenvalues lie on the unit circle in the complex plane, meaning they have an absolute value of 1.

Mathematically,

| λ | = 1 for any eigenvalue of A

This is because orthogonal matrices represent rotations and reflections, which preserve lengths and angles. This property is useful in various applications, including computer graphics and rotations in 3D space.

Eigenvalues and Similar Matrices

If A and B are two similar matrices i.e., B = P^-1AP for some invertible matrix P, then the eigenvalues of both of them remain same. Similarity transformation keeps the eigenvalues intact mathematically.

\text{det}(A - \lambda I) = \text{det}(P^{-1}AP - \lambda I) = \text{det}(P^{-1}(A - \lambda I)P) = \text{det}(A - \lambda I)

So A and B have the same characteristic polynomial as a result they share an identical eigenvalue. The central importance of this property in linear algebra is its connection with problems such as matrix equivalence and diagonalization.

Solved Examples on Properties of Eigenvalues

Example 1: Given the matrix A as follows: A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}, find the sum of its eigenvalues.

Solution:

The first step is to find the eigenvalues which could be found using characteristic equation:

\text{det}(A - \lambda I) = \text{det}\begin{pmatrix} 2 - \lambda & 1 \\ 1 & 3 - \lambda \end{pmatrix} = (2 - \lambda)(3 - \lambda) - 1 = \lambda^2 - 5\lambda + 5 = 0

The roots of the quadratic equation \lambda^2 - 5\lambda + 5 = 0 are the eigenvalues of the given matrix.

Now, using the quadratic equation roots formula:

\lambda = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}

The sum of the eigenvalues is \lambda_1 + \lambda_2 = 5, which is equal to the trace of the matrix A.

Example 2: For the matrix B given as follows: B = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}, find the product of its eigenvalues.

Solution:

As, from the properties the product of the eigenvalues is equal to the determinant of the matrix B.

So, now the determinant of matrix B is:

det(B) = (4 × 3) - (2 × 1) = 10

The product of the eigenvalues is equal to the determinant, so \lambda_1 \lambda_2 = 10.

Problem 3: Find the eigenvalues of the diagonal matrix C = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 3 \end{pmatrix}.

Solution:

Since C is a diagonal matrix so as per the properties its eigenvalues are equal to its diagonal elements.

So, now the eigenvalues are as follows:

λ₁ = 5, λ₂ = 7, λ₃ = 3.

Problem 4: Given the upper triangular matrix D: D = \begin{pmatrix} 2 & 3 \\ 0 & 4 \end{pmatrix}, find its eigenvalues.

Solution:

From the list of properties it is known that, for the upper triangular matrix, the eigenvalues are the diagonal elements.

Therefore, the eigenvalues of D are λ₁ = 2 and λ₂ = 4.

Example 5: Suppose the matrix E is given as: E = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} and its eigenvalues are λ₁ = 3 and λ₂ = -1. Find the eigenvalues of 2E.

Solution:

If λ₁ = 3 and λ₂ = -1 are eigenvalues of E, then the eigenvalues of 2E are 2λ₁ = 6 and 2λ₂ = -2.

Example 6: Find the eigenvalues of the transpose of matrix (F). Given the matrix (F) as F = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}.

Solution:

The first step is to find the eigenvalues which could be found using characteristic equation:

\text{det}(F - \lambda I) = \text{det}\begin{pmatrix} 1 - \lambda & 3 \\ 2 & 4 - \lambda \end{pmatrix} = \lambda^2 - 5\lambda - 2 = 0

Now, using the quadratic equation roots formula:

\lambda = \frac{5 \pm \sqrt{25 + 8}}{2} = \frac{5 \pm \sqrt{33}}{2}

The eigenvalues of F are \lambda_1 = \frac{5 + \sqrt{33}}{2} and \lambda_2 = \frac{5 - \sqrt{33}}{2}. Since the eigenvalues of a matrix are the same as those of its transpose, F^T has the same eigenvalues.

Example 7: Determine whether the eigenvalues of the matrix given as G = \begin{pmatrix} 2 & i \\ -i & 2 \end{pmatrix} are real or not.

Solution:

The matrix G is Hermitian as from the properties the following equation satisfies G^H = G. The characteristic equation is:

\text{det}(G - \lambda I) = \text{det}\begin{pmatrix} 2 - \lambda & i \\ -i & 2 - \lambda \end{pmatrix} = (2 - \lambda)^2 + 1 = 0

Solving for λ:

\lambda = 2 \pm 1

The eigenvalues are λ₁ = 3 and λ₂ = 1, both of which are real, by which we could say that G has real eigenvalues.

Practice Problems on Properties of Eigenvalues

Q1. Given a matrix A = \begin{pmatrix} 3 & 2 \\ 2 & 5 \end{pmatrix}, calculate the sum of its eigenvalues.

Q2. Find the product of the eigenvalues of matrix B = \begin{pmatrix} 7 & 1 \\ 3 & 4 \end{pmatrix}.

Q3. Determine the eigenvalues of the matrix C = \begin{pmatrix} 8 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \end{pmatrix}.

Q4. Find the eigenvalues of the lower triangular matrix D = \begin{pmatrix} 4 & 0 & 0 \\ 1 & 3 & 0 \\ 2 & 1 & 5 \end{pmatrix}.

Q5. If for the matrix E the eigenvalues are λ₁ = 4 and λ₂ = -2, then find the eigenvalues of 3E.

Q6. Given the matrix F = \begin{pmatrix} 2 & 5 \\ 3 & 6 \end{pmatrix}, find the eigenvalues of the transpose if the matrix F i.e. F^T.

Q7. Let G = \begin{pmatrix} 0 & 2i \\ -2i & 0 \end{pmatrix}. Are the eigenvalues of G purely imaginary or zero? Verify it by finding the eigenvalues.

Read More,

• Matrices
• Types of Matrices
• Eigenvalue and Eigenvectors

Conclusion

The properties of eigen values include the sum and product of eigenvalues, the relationships in diagonal, triangular, Hermitian, and orthogonal matrices, and the effects of matrix operations such as inversion, transposition, and scalar multiplication. The properties is necessary in various applications in mathematics, physics, engineering, and data science, which in turn supports the efficient analysis and manipulation of complex systems.